Proving the Dot Product Identity for Vector Fields and Their Curl

[Mario Carcamo]

Homework Statement

http://faculty.fiu.edu/~maxwello/phz3113/probs/set1.pdf

I'm working on problem 2. Trying to prove that the dot product between a vector field and its curl is zero.

Homework Equations

The basic identities of vector calculus and how scalar fields and vector fields interact

The Attempt at a Solution

My only real attempt is expanding what was given using an identity. What i have now is that

f(del X A) = A X del(f)

In my head the proof is trivial since by the very definition of the cross product, the new vector while be perpendicular to both vectors that served as the argument. I do have a question though. In the above notation and in the identity when they say f(del X A) do they mean to input the curl vector as an argument for the scalar function? Thanks!

 

[blue_leaf77]

In my head the proof is trivial since by the very definition of the cross product, the new vector while be perpendicular to both vectors that served as the argument.

I don't precisely know which vectors you were referring to by "both vectors", but I have a feeling that you are concerned with $\mathbf{A}$ being always perpendicular to $\nabla \times \mathbf{A}$ because the cross product in between the last expression makes it perpendicular to $\mathbf{A}$. I believe this is not necessarily the case. Anyway, the more elegant way is to multiply by dot product the identity you have there with $\mathbf{A}$.

In the above notation and in the identity when they say f(del X A) do they mean to input the curl vector as an argument for the scalar function? Thanks!

No, they are just vector multiplied with a scalar.

 

[Mario Carcamo]

I don't precisely know which vectors you were referring to by "both vectors", but I have a feeling that you are concerned with $\mathbf{A}$ being always perpendicular to $\nabla \times \mathbf{A}$ because the cross product in between the last expression makes it perpendicular to $\mathbf{A}$. I believe this is not necessarily the case. Anyway, the more elegant way is to multiply by dot product the identity you have there with $\mathbf{A}$.

No, they are just vector multiplied with a scalar.

Yeah what i mean is if A X B = C then C must be perpendicular to A and B so if your doing Del X A = C then C must be perpendicular to A but maybe that is not the case when your talking about del?

 

[blue_leaf77]

but maybe that is not the case when your talking about del?

One of the identities involving curl is $\nabla \times (f\mathbf{A}) + \mathbf{A}\times \nabla f= f(\nabla \times \mathbf{A})$. Try multiplying by dot product both sides with $\mathbf{A}$. An example in electromagnetism is one of the Maxwell equations, $\nabla \times \mathbf{E} = -\partial \mathbf{B} /\partial t$, in most cases electric and magnetic fields are perpendicular, except for certain cases like that in the wave propagation in waveguides (for example see http://physics.stackexchange.com/qu...ric-and-magnetic-fields-are-not-perpendicular).

 

[Mario Carcamo]

One of the identities involving curl is $\nabla \times (f\mathbf{A}) + \mathbf{A}\times \nabla f= f(\nabla \times \mathbf{A})$. Try multiplying by dot product both sides with $\mathbf{A}$. An example in electromagnetism is one of the Maxwell equations, $\nabla \times \mathbf{E} = -\partial \mathbf{B} /\partial t$, in most cases electric and magnetic fields are perpendicular, except for certain cases like that in the wave propagation in waveguides (for example see http://physics.stackexchange.com/qu...ric-and-magnetic-fields-are-not-perpendicular).

yeah i got the answer thanks alot!